How large a sample should be taken if the population mean is to be estimated with 99% confidence to within $75? The population has a standard deviation of $900.
Thanks to Dr Bob for presenting this solution earlier : )
mu = bar x +/- 3(sigma/sqrt N)
mu is the population mean.
bar x is the average.
3 gives the 99% confidence interval.
sigma is standard deviation. N is the number of sample. Solve for N.
To determine the sample size required to estimate the population mean with a specific level of confidence and margin of error, we can use the formula:
n = (Z * σ / E)^2
where:
n = required sample size
Z = Z-score corresponding to the desired level of confidence.
σ = population standard deviation
E = desired margin of error
In this case, we want a 99% confidence level, which corresponds to a Z-score of 2.576 (consulting a Z-table or using a calculator).
Substituting the given values into the formula:
n = (2.576 * 900 / 75)^2
Calculating this expression:
n = (2318.4)^2
n ≈ 5,366,023
Therefore, the sample size required to estimate the population mean with a 99% confidence level and within a margin of error of $75 is approximately 5,366,023. However, since sample sizes are typically not fractions or decimals, you would need to round up to the nearest whole number. Thus, a sample size of at least 5,366,023 would be necessary.